大家好，又见面了，我是全栈君，今天给大家准备了Idea注册码。
Given a sequence of five distinct times written in the format hh : mm , where hh are two digits representing full hours (00 <= hh <= 23) and mm are two digits representing minutes (00 <= mm <= 59) , you are to write a program that finds the median, that is, the third element of the sorted sequence of times in a nondecreasing order of their associated angles. Ties are broken in such a way that an earlier time precedes a later time.
For example, suppose you are given a sequence (06:05, 07:10, 03:00, 21:00, 12:55) of times. Because the sorted sequence is (12:55, 03:00, 21:00, 06:05, 07:10), you are to report 21:00.
Input
The input consists of T test cases. The number of test cases (T) is given on the first line of the input file. Each test case is given on a single line, which contains a sequence of five distinct times, where times are given in the format hh : mm and are separated by a single space.
Output
Print exactly one line for each test case. The line is to contain the median in the format hh : mm of the times given. The following shows sample input and output for three test cases.
Sample Input
3 00:00 01:00 02:00 03:00 04:00 06:05 07:10 03:00 21:00 12:55 11:05 12:05 13:05 14:05 15:05
Sample Output
02:00 21:00 14:05
Source:
Asia 2003, Seoul (South Korea)
题意：
//给出 5 个时刻，按时钟的时针，分针夹角从小到大排序，
//输出中间的时刻。
代码例如以下：
#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
struct TIME
{
int h;
int m;
int angle;
}a[7];
int cal(TIME TT)
{
if(TT.h > 12)
{
TT.h-=12;
}
int tt = abs((TT.h*60 + TT.m) - TT.m*12);
//原式为:TT.h*30+(TT.m/60)*30-a.m*6;
if(tt > 360)
tt = 720 - tt;
return tt;
}
bool cmp(TIME A, TIME B)
{
if(A.angle != B.angle)
{
return A.angle < B.angle;
}
else if(A.h != B.h)
{
return A.h < B.h;
}
else
return A.m < B.m;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
for(int i = 0; i < 5; i++)
{
scanf("%d:%d",&a[i].h,&a[i].m);
a[i].angle = cal(a[i]);
}
sort(a,a+5,cmp);
printf("%02d:%02d\n",a[2].h,a[2].m);
}
return 0;
}
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